CSE 143 Lecture 16

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# CSE 143 Lecture 16 - PowerPoint PPT Presentation

CSE 143 Lecture 16. Recursive Backtracking reading: &quot;Appendix R&quot; on course web site slides created by Marty Stepp and Hélène Martin http://www.cs.washington.edu/143/ ideas and examples taken from Stanford University CS slides/lectures. Exercise: Dice rolls.

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### CSE 143Lecture 16

Recursive Backtracking

reading: "Appendix R" on course web site

slides created by Marty Stepp and Hélène Martin

http://www.cs.washington.edu/143/

ideas and examples taken from Stanford University CS slides/lectures

Exercise: Dice rolls
• Write a method diceRoll that accepts an integer parameter representing a number of 6-sided dice to roll, and output all possible combinations of values that could appear on the dice.

diceRoll(2); diceRoll(3);

[1, 1, 1]

[1, 1, 2]

[1, 1, 3]

[1, 1, 4]

[1, 1, 5]

[1, 1, 6]

[1, 2, 1]

[1, 2, 2]

...

[6, 6, 4]

[6, 6, 5]

[6, 6, 6]

Examining the problem
• We want to generate all possible sequences of values.

for (each possible first die value):

for (each possible second die value):

for (each possible third die value):

...

print!

• This is called a depth-first search
• How can we completely explore such a large search space?
Backtracking
• backtracking: Finding solution(s) by trying partial solutions and then abandoning them if they are not suitable.
• a "brute force" algorithmic technique (tries all paths)
• often implemented recursively

Applications:

• producing all permutations of a set of values
• parsing languages
• games: anagrams, crosswords, word jumbles, 8 queens
• combinatorics and logic programming
Backtracking algorithms

A general pseudo-code algorithm for backtracking problems:

Explore(choices):

• if there are no more choices to make: stop.
• else, for each available choice C:
• Make choice C.
• Explore the remaining choices.
• Un-make choice C, if necessary. (backtrack!)
A decision tree

...

...

...

...

...

...

...

...

Private helpers
• Often the method doesn't accept the parameters you want.
• So write a private helper that accepts more parameters.
• Extra params can represent current state, choices made, etc.

public int methodName(params):

...

return helper(params, moreParams);

private int helper(params, moreParams):

...

(use moreParams to help solve the problem)

Exercise solution

// Prints all possible outcomes of rolling the given

// number of six-sided dice in [#, #, #] format.

public static void diceRolls(int dice) {

List<Integer> chosen = new ArrayList<Integer>();

diceRolls(dice, chosen);

}

// private recursive helper to implement diceRolls logic

private static void diceRolls(int dice,

List<Integer> chosen) {

if (dice == 0) {

System.out.println(chosen); // base case

} else {

for (int i = 1; i <= 6; i++) {

diceRolls(dice - 1, chosen); // explore

chosen.remove(chosen.size() - 1); // un-choose

}

}

}

Exercise: Dice roll sum
• Write a method diceSum similar to diceRoll, but it also accepts a desired sum and prints only combinations that add up to exactly that sum.

diceSum(2, 7); diceSum(3, 7);

[1, 6]

[2, 5]

[3, 4]

[4, 3]

[5, 2]

[6, 1]

[1, 1, 5]

[1, 2, 4]

[1, 3, 3]

[1, 4, 2]

[1, 5, 1]

[2, 1, 4]

[2, 2, 3]

[2, 3, 2]

[2, 4, 1]

[3, 1, 3]

[3, 2, 2]

[3, 3, 1]

[4, 1, 2]

[4, 2, 1]

[5, 1, 1]

Optimizations
• We need not visit every branch of the decision tree.
• Some branches are clearly not going to lead to success.
• We can preemptively stop, or prune, these branches.
• Inefficiencies in our dice sum algorithm:
• Sometimes the current sum is already too high.
• (Even rolling 1 for all remaining dice would exceed the desired sum.)
• Sometimes the current sum is already too low.
• (Even rolling 6 for all remaining dice would exceed the desired sum.)
• When finished, the code must compute the sum every time.
• (1+1+1 = ..., 1+1+2 = ..., 1+1+3 = ..., 1+1+4 = ..., ...)
Exercise solution, improved

public static void diceSum(int dice, int desiredSum) {

List<Integer> chosen = new ArrayList<Integer>();

diceSum2(dice, desiredSum, chosen, 0);

}

private static void diceSum(int dice, int desiredSum,

List<Integer> chosen, int sumSoFar) {

if (dice == 0) {

if (sumSoFar == desiredSum) {

System.out.println(chosen);

}

} else if (sumSoFar <= desiredSum &&

sumSoFar + 6 * dice >= desiredSum) {

for (int i = 1; i <= 6; i++) {

diceSum(dice - 1, desiredSum, chosen, sumSoFar + i);

chosen.remove(chosen.size() - 1);

}

}

}

Backtracking strategies
• When solving a backtracking problem, ask these questions:
• What are the "choices" in this problem?
• What is the "base case"? (How do I know when I'm out of choices?)
• How do I "make" a choice?
• Do I need to create additional variables to remember my choices?
• Do I need to modify the values of existing variables?
• How do I explore the rest of the choices?
• Do I need to remove the made choice from the list of choices?
• Once I'm done exploring, what should I do?
• How do I "un-make" a choice?
Exercise: Permutations
• Write a method permute that accepts a string as a parameter and outputs all possible rearrangements of the letters in that string. The arrangements may be output in any order.
• Example:permute("TEAM")outputs the followingsequence of lines:
Examining the problem
• We want to generate all possible sequences of letters.

for (each possible first letter):

for (each possible second letter):

for (each possible third letter):

...

print!

• Each permutation is a set of choices or decisions:
• Which character do I want to place first?
• Which character do I want to place second?
• ...
• solution space: set of all possible sets of decisions to explore
Exercise solution

// Outputs all permutations of the given string.

public static void permute(String s) {

permute(s, "");

}

private static void permute(String s, String chosen) {

if (s.length() == 0) {

// base case: no choices left to be made

System.out.println(chosen);

} else {

// recursive case: choose each possible next letter

for (int i = 0; i < s.length(); i++) {

char c = s.charAt(i); // choose

s = s.substring(0, i) + s.substring(i + 1);

chosen += c;

permute(s, chosen); // explore

s = s.substring(0, i) + c + s.substring(i + 1);

chosen = chosen.substring(0, chosen.length() - 1);

} // un-choose

}

}

Exercise solution 2

// Outputs all permutations of the given string.

public static void permute(String s) {

permute(s, "");

}

private static void permute(String s, String chosen) {

if (s.length() == 0) {

// base case: no choices left to be made

System.out.println(chosen);

} else {

// recursive case: choose each possible next letter

for (int i = 0; i < s.length(); i++) {

String ch = s.substring(i, i + 1); // choose

String rest = s.substring(0, i) + // remove

s.substring(i + 1);

permute(rest, chosen + ch); // explore

}

} // (don't need to "un-choose" because

} // we used temp variables)

Exercise: Combinations
• Write a method combinations that accepts a string s and an integer k as parameters and outputs all possible k -letter words that can be formed from unique letters in that string. The arrangements may be output in any order.
• Example:combinations("GOOGLE", 3)outputs the sequence oflines at right.
• To simplify the problem, you may assumethat the string s contains at least kunique characters.
Initial attempt

public static void combinations(String s, int length) {

combinations(s, "", length);

}

private static void combinations(String s, String chosen, int length) {

if (length == 0) {

System.out.println(chosen); // base case: no choices left

} else {

for (int i = 0; i < s.length(); i++) {

String ch = s.substring(i, i + 1);

if (!chosen.contains(ch)) {

String rest = s.substring(0, i) + s.substring(i + 1);

combinations(rest, chosen + ch, length - 1);

}

}

}

}

• Problem: Prints same string multiple times.
Exercise solution

public static void combinations(String s, int length) {

Set<String> all = new TreeSet<String>();

combinations(s, "", all, length);

for (String comb : all) {

System.out.println(comb);

}

}

private static void combinations(String s, String chosen,

Set<String> all, int length) {

if (length == 0) {

all.add(chosen); // base case: no choices left

} else {

for (int i = 0; i < s.length(); i++) {

String ch = s.substring(i, i + 1);

if (!chosen.contains(ch)) {

String rest = s.substring(0, i) + s.substring(i + 1);

combinations(rest, chosen + ch, all, length - 1);

}

}

}

}